Undeclared Variable mysqli - php

Question

I'm having trouble with mysqli and prepared statements. I've just started learning mysqli an hour and am having trouble not understanding why I'm getting these two errors:
Notice: Undefined variable: mysqli in /opt/lampp/htdocs/lr/testingi.php on line 17
Fatal error: Call to a member function prepare() on a non-object in /opt/lampp/htdocs
/lr/testingi.php on line 17
I have a file that contains the database connection. Here it is.
$mysqli = new mysqli("localhost", "user", "password", "db");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
Here is the test file that is reproducing the error.
session_start();
require_once 'core/database/connect.php';
function user_id_from_username ($username) {
if ($stmt = $mysqli->prepare("SELECT `user_id` FROM `users` WHERE `username` = ?"))
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->bind_result($user_id);
echo $user_id;
$stmt->close();
}
$username = 'Jason';
user_id_from_username ($username);

Answer 1

Looks like you're not passing $mysqli into the user_id_from_username function.
2 quick options:
1. A global
function user_id_from_username ($username) {
global $mysqli;
if ($stmt = $mysqli->prepare("SELECT `user_id` FROM `users` WHERE `username` = ?"))
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->bind_result($user_id);
echo $user_id;
$stmt->close();
}
2. a second parameter
function user_id_from_username ($mysqli, $username) {//..}
user_id_from_username($mysqli, $username);